## Permutations and Combinations

### 2.9 Extra: Fun with Poker Hands

One plays poker with a deck of 52 cards, which come in 4 suits (hearts, clubs, spades, diamonds) with 13 values per suit (A, 2, 3, …, 10, J, Q, K).

In poker one is dealt five cards and certain combinations of cards are deemed valuable. For example, a “four of a kind” consists of four cards of the same value and a fifth card of arbitrary value. A “full house” is a set of three cards of one value and two cards of a second value. A “flush” is a set of five cards of the same suit. The order in which one holds the cards in ones hand is immaterial.

 EXAMPLE: How many flushes are possible in poker?

Answer: Again this is a multi-stage problem with each stage being its own separate labeling problem. One way to help tease apart stages is to image that you’ve been given the task of writing a computer program to create poker hands. How will you instruct the computer to create a flush?

First of all, there are four suits – hearts, spades, clubs and diamonds – and we need to choose one to use for our flush. That is, we need to label one suit as “used” and three suits as “not used.” There are $$\dfrac{4!}{1!3!} = 4$$ ways to do this.

Second stage: Now that we have a suit, we need to choose five cards from the 13 cards of that suit to use for our hand. Again, this is a labeling problem  – label five cards as “used” and eight cards as “not used.” There are $$\dfrac{13!}{5!8!} = 1287$$ ways to do this.

By the multiplication principle there are $$4 \times 1287 = 5148$$  ways to compete both stages. That is, there are $$5148$$ possible flushes.

COMMENT: There are $$\dfrac{52!}{5!47!} = 2598960$$  five-card hands in total in poker. (Why?) The chances of being dealt a flush are thus: $$\dfrac{5148}{2598960} \approx 0.20\%$$.

 EXAMPLE: How many full houses are possible in poker?

Answer: This problem is really a three-stage labeling issue.

First we must select which of the thirteen card values – A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K –  is going to be used for the triple, which will be used for the double, and which 11 values are going to be ignored. There are $$\dfrac{13!}{1!1!11!} = 13 \times 12 = 156$$ ways to accomplish this task.

Among the four cards of the value selected for the triple, three will be used for the triple and one will be ignored. There are $$\dfrac{4!}{3!1!} = 4$$ ways to accomplish this task. Among the four cards of the value selected for the double, two will be used and two will be ignored. There are $$\dfrac{4!}{2!2!} = 6$$ ways to accomplish this.

By the multiplication principle, there are $$156 \times 4 \times 6 = 3744$$  possible full houses.

COMMENT: High-school teacher Sam Miskin recently used this labeling method to count poker hands with his high-school students. To count how many “one pair hands” (that is, hands with one pair of cards the same numerical value and three remaining cards each of different value) he found it instructive bring 13 students to the front of the room and hand each student four cards the same value from a single deck of cards.

He then asked the remaining students to select which of the thirteen students should be the “pair” and which three should be the “singles.” He had the remaining nine students return to their seats.

He then asked the “pair” student to raise his four cards in the air and asked the seated students to select which two of the four should be used for the pair. He then asked each of the three “single” students in turn to hold up their cards while the seated students selected on one the four cards to make a singleton.

This process made the multi-stage procedure clear to all and the count of possible one pair hands, namely,

$$\dfrac{13!}{1!3!9!} \times \dfrac{4!}{2!2!} \times 4 \times 4 \times 4$$

 Exercise: “Two pair” consists of two cards of one value, two cards of a different value, and a third card of a third value. What are the chances of being dealt two-pair in poker?

 EXAMPLE: A “straight” consists of five cards with values forming a string of five consecutive values (with no “wrap around”). For example, 45678, A2345 and 10JQKA are considered straights, but KQA23 is not. (Suits are immaterial for straights.)  How many different straights are there in poker?

Answer: A straight can begin with A, 2, 3, 4, 5, 6, 7, 8, 9 or 10. We must first select which of these values is to be the start of our straight. There are 10 choices.

For the starting value we must select which of the four suits it will be. There are 4 choices.

There are also 4 choices for the suit of the second card in the straight, 4 for the third, 4 for the fourth, and 4 for the fifth.

By the multiplication principle, the total number of straights is:

$$10 \times 4 \times 4 \times 4 \times 4 \times 4 = 10240$$.

The chances of being dealt a straight are about 0.39%.

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